Answer:
For a rectangle of length L and width W, the area is:
A = L*W
In this case we have:
L = x + 2
w = x
Then the area of this rectangle is:
A = (x + 2)*x = x^2 + 2*x
And we know that this area is equal to 168 square units, then:
168 = x^2 + 2*x
We can rewrite this as:
x^2 + 2*x - 168 = 0
This is a quadratic equation, and the solutions are given by Bhaskara's equation, such that for an equation of the form:
a*x^2 + b*x + c = 0
The solutions are given by:
[tex]x = \frac{-b \pm\sqrt{b^2 - 4*a*c} }{2*a}[/tex]
In our case, the solutions are:
[tex]x = \frac{-2 \pm \sqrt{2^2 - 4*1*(-168)} }{2} = \frac{-2 \pm 26}{2}[/tex]
Then the two solutions are:
x = (2 - 26)/2 = -12
x = (2 + 26)/2 = 14
Notice that x is the width of the rectangle, and we can not have a negative width, so we can discard the first option.
Then we can conclude that:
x = 14
Then the width of the rectangle is 14 units long, and the length is 14 + 2 = 16 units long.
Now, how the lengths are related to eachother?
The length is 2 units longer than the width.
